Can we cover the entire plane with the square with area 1/n for each positive integer n?

Question

We have one square with area 1/n for each positive integer n.

Is it possible to place these squares in the xy-plane in such a way that they completely cover the entire plane. If Yes, can you describe how this can be done (you might also want to draw a picture). If No, explain why this cannot be done.

The sum of their areas correspond to the harmonic series which is divergent. That is the 'total' area is 'infinite'. The 'total' area of the entire plane is also 'infinite'. Yet, infinity has some levels, how can we compare them?

Edit: We could consider both cases :

the first case : overlap of squares is allowed

the second case : overlap of squares is not allowed. I am particularly interested in that case.

Answer 1

The argument of @user1952009 can be improved in order to give a tiling. Since the harmonic series is divergent, we may fill $\mathbb{R}^+\times \mathbb{R}^+$ with some of the squares whose area is $\frac{1}{4k+1}$. We may fill $\mathbb{R}^+\times \mathbb{R}^+$ also with some of the squares whose area is $\frac{1}{4k+2}$, or $\frac{1}{4k+3}$, or $\frac{1}{4k+4}$. So we may fill $\mathbb{R}^2$ by using some of the given squares, by just glueing the four copies of $\mathbb{R}^+\times\mathbb{R}^+$, each of them tiled with some squares with area $\frac{1}{4k+j}$, for $j\in\{1,2,3,4\}$. Assume that the largest unused square, $Q_1$, has area $\frac{1}{n_1}$.

By the same argument as above we may fill $\mathbb{R}^2\setminus Q_1$ by using exactly the same squares as before, then place $Q_1$ in the hole. By moving a bit those four corners we may make room for the other unused squares. So it is possible to tile $\mathbb{R}^2$ with squares having areas $\frac{1}{1},\frac{1}{2},\frac{1}{3},\ldots$, each of them used once, but the tiling procedure is quite non-constructive. Also because how we move the four corners above depends on the unused squared having a finite total area or not.

Answer 2

I'm not quite familiar with real analysis, after all, my major is in physics. So forgive me for some tiny mistakes, but I believe the idea is right.

Lemma. When you have finitely many squares with their total area to be 3. Then you can cover a unit square with them parallelly. By parallelly we mean each edge of the individual squares to be parallel to the unit square.

Proof:Denote the unit square by $M$. Firstly, let us sort the small squares from large to small parallelly along the bottom side of $M$,until the sum of the edges to be no less than $1$. Suppose the edge of the last square is $h_i$.

Accordingly, let us perform the same operation. This time, we sort the rest of the squares just above the first line of squares, where the bottom edges are right $h_1$ above $M$. Suppose the edge of the last square is $h_2$.

After several operations stated above. we have a series $h_i$. and the largest edge of the $(i+1) th$ line is less than $h_1$. and the largest edge of the first square is less than $1$.

We have: $$ Sum\ of\ the\ squares \leq 1\times (1+h_1) +h_1\times (1+h_2)+...\leq 1+2h_1+2h_2+... $$

That is to say: $$ 1+2h_1+2h_2+...\geq 3 $$

Which implies $h_1+h_2+...\geq 1$.

Back to the problem:

We have: $$ 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}\geq 1+ \frac{1}{2} + 2\times \frac{1}{4} + 4\times \frac{1}{8} + 8\times \frac{1}{16}=3\\ ... $$ so we take the first 16 squares out the cover a unit square.

Accordingly we can take the squares of area $\frac{1}{2^{6k+4}+1}...\frac{1}{2^{6k+10}}$ to cover a unit square.

Let us now construct a map form $k$ to $(x,y)$, where $x,y$ are integers. Then let the $k th$ unit square to be centered at $(x,y)$, thus complete coverage.

PS. I'm not a native English user, hope you can forgive my poor English.